# After how many hours does a quantity becomes less than 1% initial quantity?

Life of substance reduces to half at the end of one hour i.e its quantity reduces to one half of what it was at the beginning of one hour .

In how many hours , the quantity becomes less than $1$% initial quantity..

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After $1$ hour, we have $\frac{1}{2}$ left.

After $2$ hours, we have one-half of $\frac{1}{2}$ left, so $\frac{1}{4}$.

After $3$ hours, we have one-half of $\frac{1}{4}$ left, so $\frac{1}{8}$.

After $4$ hours, we have one-half of $\frac{1}{8}$ left, so $\frac{1}{16}$.

Continue. After $6$ hours, we have $\frac{1}{64}$ left, after $7$, we have $\frac{1}{128}$. Now we are below $1\%$.

If the looked for answer is an integer, that integer is $7$. But if the decay process takes place continuously, as it probably does, the answer will be a number between $6$ and $7$. To find that number, note that if at the beginning we have an amount $A$, then after $t$ hours we have an amount $A(t)$, where $$A(t)=\frac{a}{2^t}.$$

We want the time $t$ until $A$ decays to $\frac{A}{100}$. So we have $$\frac{A}{100}=\frac{A}{2^t}.$$ This simplifies to $2^t=100$.

To solve this equation for $t$, we can use our calculator to hunt and peck our way. For example, my calculator says that $2^{6.5}\approx 90.51$, so $t\gt 6.5$. Soon you can zoom in on an excellent approximation.

Or else we can take logarithms, to any base you like. We get $\log(2^t)=\log(100)$. So $$t\log 2=\log(100),$$ and therefore $$t=\frac{\log(100)}{\log 2}.$$ I get $t\approx 6.644$.

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We use the half life formula to solve. Assume that you start with $100$% of the quantity. You want to find the time it takes to reach $1$%, so let us set up an equation.

$$.01 = \exp(-kt).$$

Notice I have an unknown value in this equation ($k$). We can find this by using what we are given,

$$1/2 = \exp(-k)$$

Here we have just $\exp(-k)$ since we know $t=1\Rightarrow 1/2$ quantity left. Now use logarithms to solve for $k$ and then solve the first equation I posted.

Since $\ln(1/2)=-\ln(2)$ we have $k=\ln(2)$. Now solve

$$.01 = \exp(-\ln(2)t).$$

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is there any simple method to solve this question ? \ – SSK Jun 13 '13 at 18:45
@SSK Not if the exact answer is called for. Many people in the physical sciences use the "rule-of-thumb" that seven half-lives reduce the initial amount of a (radioactive or biological) substance to about 1% , since $\ ( \frac{1}{2} )^7 = \frac{1}{128} < \frac{1}{100} \ .$ (On the other hand, a lot of physicists use "e-folding times", for which the quantity falls to under 1% of its initial value in five e-folding times, since $\ ( \frac{1}{e} )^5 \approx \frac{1}{148} < \frac{1}{100} \ .$ ) – RecklessReckoner Jun 13 '13 at 18:51
@amWhy I think a continuous process is intended: the wording in OP's statement might suggest otherwise, but I think that is just the way OP expressed the situation. – RecklessReckoner Jun 13 '13 at 18:53

Let be $n$ the initial quantity. If at each hour the quantity reduces to half of what was at the beguining of this hour, one can say that $\displaystyle \frac{n}{2^x}$ gives the quantity at the end of each $x$ hour.

If $\displaystyle \frac{n}{2^x}$ gives the quantity at each hour, then $\displaystyle n \cdot \frac{2^x}{n}=\frac{1}{2^x}$ gives the relative percentage. And so, one needs to solve $\frac{1}{2^x}\leq \frac{1}{100}$.

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